Why is $\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}$?

Question

Currently I have a problem to understand why $$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}\,?$$

My calculations get me:

$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}\equiv \frac{(1+\cos x)^2}{1+\sin^2x}.$$

Why is this not correct? We have a rational number in the denumerator. Thus the numerator of this rational number should move above. The other formula did this too. But why is there $2\cos x+\cos^2x$?

Thanks in advance! Stay healthy!

Answer 1

If you multiply the numerator and the denominator by $(1+\cos x)^2$, you get:

$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2}\times\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2\left(1+(\frac{-\sin x}{1+\cos x})^2\right)}\\=\frac{(1+\cos x)^2}{(1+\cos x)^2+\sin^2x}$$